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%Tutorial slides on Python.

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% Author: FOSSEE

% Copyright (c) 2009, FOSSEE, IIT Bombay

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% Title page

\title[Solving Equations \& ODEs]{Python for Science and Engg:\\SciPy}

\author[FOSSEE] {FOSSEE}

\institute[IIT Bombay] {Department of Aerospace Engineering\\IIT Bombay}

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    \frametitle{Outline}

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    \frametitle{Outline}

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\begin{document}

\begin{frame}

  \maketitle

\end{frame}

%% \begin{frame}

%%   \frametitle{Outline}

%%   \tableofcontents

%%   % You might wish to add the option [pausesections]

%% \end{frame}

\section{Least Squares Fit}

\begin{frame}[fragile]

\frametitle{$L$ vs. $T^2$ - Scatter}

Linear trend visible.

\vspace{-0.1in}

\begin{figure}

\includegraphics[width=4in]{data/L-Tsq-points}

\end{figure}

\end{frame}

\begin{frame}[fragile]

\frametitle{$L$ vs. $T^2$ - Line}

This line does not make any mathematical sense.

\vspace{-0.1in}

\begin{figure}

\includegraphics[width=4in]{data/L-Tsq-Line}

\end{figure}

\end{frame}

\begin{frame}[fragile]

\frametitle{$L$ vs. $T^2$ - Least Square Fit}

This is what our intention is.

\vspace{-0.1in}

\begin{figure}

\includegraphics[width=4in]{data/least-sq-fit}

\end{figure}

\end{frame}

\begin{frame}[fragile]

\frametitle{Matrix Formulation}

\begin{itemize}

\item We need to fit a line through points for the equation $T^2 = m \cdot L+c$

\item In matrix form, the equation can be represented as $T_{sq} = A \cdot p$, where $T_{sq}$ is

  $\begin{bmatrix}

  T^2_1 \\

  T^2_2 \\

  \vdots\\

  T^2_N \\

  \end{bmatrix}$

, A is   

  $\begin{bmatrix}

  L_1 & 1 \\

  L_2 & 1 \\

  \vdots & \vdots\\

  L_N & 1 \\

  \end{bmatrix}$

  and p is 

  $\begin{bmatrix}

  m\\

  c\\

  \end{bmatrix}$

\item We need to find $p$ to plot the line

\end{itemize}

\end{frame}

\begin{frame}[fragile]

\frametitle{Getting $L$ and $T^2$}

%If you \alert{closed} IPython after session 2

\begin{lstlisting}

In []: L, T = loadtxt('pendulum.txt', 

                      unpack=True)

In []: tsq = T*T

\end{lstlisting}

\end{frame}

\begin{frame}[fragile]

\frametitle{Generating $A$}

\begin{lstlisting}

In []: A = array([L, ones_like(L)])

In []: A = A.T

\end{lstlisting}

\end{frame}

\begin{frame}[fragile]

\frametitle{\typ{lstsq} \ldots}

\begin{itemize}

\item Now use the \typ{lstsq} function

\item Along with a lot of things, it returns the least squares solution

\end{itemize}

\begin{lstlisting}

In []: result = lstsq(A,tsq)

In []: coef = result[0]

\end{lstlisting}

\end{frame}

\begin{frame}[fragile]

\frametitle{Least Square Fit Line \ldots}

We get the points of the line from \typ{coef}

\begin{lstlisting}

In []: Tline = coef[0]*L + coef[1]

In []: Tline.shape

\end{lstlisting}

\begin{itemize}

\item Now plot \typ{Tline} vs. \typ{L}, to get the Least squares fit line. 

\end{itemize}

\begin{lstlisting}

In []: plot(L, Tline, 'r')

In []: plot(L, tsq, 'o')

\end{lstlisting}

\end{frame}

\begin{frame}[fragile]

\frametitle{Least Squares Fit}

\vspace{-0.15in}

\begin{figure}

\includegraphics[width=4in]{data/least-sq-fit}

\end{figure}

\end{frame}

\section{Solving linear equations}

\begin{frame}[fragile]

\frametitle{Solution of equations}

Consider,

  \begin{align*}

    3x + 2y - z  & = 1 \\

    2x - 2y + 4z  & = -2 \\

    -x + \frac{1}{2}y -z & = 0

  \end{align*}

Solution:

  \begin{align*}

    x & = 1 \\

    y & = -2 \\

    z & = -2

  \end{align*}

\end{frame}

\begin{frame}[fragile]

\frametitle{Solving using Matrices}

Let us now look at how to solve this using \kwrd{matrices}

  \begin{lstlisting}

In []: A = array([[3,2,-1],

                  [2,-2,4],                   

                  [-1, 0.5, -1]])

In []: b = array([1, -2, 0])

In []: x = solve(A, b)

  \end{lstlisting}

\end{frame}

\begin{frame}[fragile]

\frametitle{Solution:}

\begin{lstlisting}

In []: x

Out[]: array([ 1., -2., -2.])

\end{lstlisting}

\end{frame}

\begin{frame}[fragile]

\frametitle{Let's check!}

\begin{small}

\begin{lstlisting}

In []: Ax = dot(A, x)

In []: Ax

Out[]: array([  1.00000000e+00,  -2.00000000e+00,  -1.11022302e-16])

\end{lstlisting}

\end{small}

\begin{block}{}

The last term in the matrix is actually \alert{0}!\\

We can use \kwrd{allclose()} to check.

\end{block}

\begin{lstlisting}

In []: allclose(Ax, b)

Out[]: True

\end{lstlisting}

\inctime{10}

\end{frame}

\begin{frame}[fragile]

\frametitle{Problem}

Solve the set of equations:

\begin{align*}

  x + y + 2z -w & = 3\\

  2x + 5y - z - 9w & = -3\\

  2x + y -z + 3w & = -11 \\

  x - 3y + 2z + 7w & = -5\\

\end{align*}

\inctime{5}

\end{frame}

\begin{frame}[fragile]

\frametitle{Solution}

Use \kwrd{solve()}

\begin{align*}

  x & = -5\\

  y & = 2\\

  z & = 3\\

  w & = 0\\

\end{align*}

\end{frame}

\section{Finding Roots}

\begin{frame}[fragile]

\frametitle{SciPy: \typ{roots}}

\begin{itemize}

\item Calculates the roots of polynomials

\item To calculate the roots of $x^2-5x+6$ 

\end{itemize}

\begin{lstlisting}

  In []: coeffs = [1, -5, 6]

  In []: roots(coeffs)

  Out[]: array([3., 2.])

\end{lstlisting}

\vspace*{-.2in}

\begin{center}

\includegraphics[height=1.6in, interpolate=true]{data/roots}    

\end{center}

\end{frame}

\begin{frame}[fragile]

\frametitle{SciPy: \typ{fsolve}}

\begin{small}

\begin{lstlisting}

  In []: from scipy.optimize import fsolve

\end{lstlisting}

\end{small}

\begin{itemize}

\item Finds the roots of a system of non-linear equations

\item Input arguments - Function and initial estimate

\item Returns the solution

\end{itemize}

\end{frame}

\begin{frame}[fragile]

\frametitle{\typ{fsolve}}

Find the root of $sin(z)+cos^2(z)$ nearest to $0$

\vspace{-0.1in}

\begin{center}

\includegraphics[height=2.8in, interpolate=true]{data/fsolve}    

\end{center}

\end{frame}

\begin{frame}[fragile]

\frametitle{\typ{fsolve}}

Root of $sin(z)+cos^2(z)$ nearest to $0$

\begin{lstlisting}

In []: fsolve(sin(z)+cos(z)*cos(z), 0)

NameError: name 'z' is not defined

\end{lstlisting}

\end{frame}

\begin{frame}[fragile]

\frametitle{\typ{fsolve}}

\begin{lstlisting}

In []: z = linspace(-pi, pi)

In []: fsolve(sin(z)+cos(z)*cos(z), 0)

\end{lstlisting}

\begin{small}

\alert{\typ{TypeError:}}

\typ{'numpy.ndarray' object is not callable}

\end{small}

\end{frame}

\begin{frame}[fragile]

\frametitle{Functions - Definition}

We have been using them all along. Now let's see how to define them.

\begin{lstlisting}

In []: def g(z):

 ....:     return sin(z)+cos(z)*cos(z)

\end{lstlisting}

\begin{itemize}

\item \typ{def} -- keyword

\item name: \typ{g}

\item arguments: \typ{z}

\item \typ{return} -- keyword

\end{itemize}

\end{frame}

\begin{frame}[fragile]

\frametitle{Functions - Calling them}

\begin{lstlisting}

In []: g()

---------------------------------------

\end{lstlisting}

\alert{\typ{TypeError:}}\typ{g() takes exactly 1 argument}

\typ{(0 given)}

\begin{lstlisting}

In []: g(0)

Out[]: 1.0

In []: g(1)

Out[]: 1.1333975665343254

\end{lstlisting}

More on Functions later \ldots

\end{frame}

\begin{frame}[fragile]

\frametitle{\typ{fsolve} \ldots}

Find the root of $sin(z)+cos^2(z)$ nearest to $0$

\begin{lstlisting}

In []: fsolve(g, 0)

Out[]: -0.66623943249251527

\end{lstlisting}

\begin{center}

\includegraphics[height=2in, interpolate=true]{data/fsolve}    

\end{center}

\end{frame}

\begin{frame}[fragile]

  \frametitle{Exercise Problem}

  Find the root of the equation $x^2 - sin(x) + cos^2(x) = tan(x)$ nearest to $0$

\end{frame}

\begin{frame}[fragile]

  \frametitle{Solution}

  \begin{small}

  \begin{lstlisting}

def g(x):

    return x**2 - sin(x) + cos(x)*cos(x) - tan(x)

fsolve(g, 0)

  \end{lstlisting}

  \end{small}

  \begin{center}

\includegraphics[height=2in, interpolate=true]{data/fsolve_tanx}

  \end{center}

\end{frame}

%% \begin{frame}[fragile]

%% \frametitle{Scipy Methods \dots}

%% \begin{small}

%% \begin{lstlisting}

%% In []: from scipy.optimize import fixed_point

%% In []: from scipy.optimize import bisect

%% In []: from scipy.optimize import newton

%% \end{lstlisting}

%% \end{small}

%% \end{frame}

\section{ODEs}

\begin{frame}[fragile]

\frametitle{Solving ODEs using SciPy}

\begin{itemize}

\item Consider the spread of an epidemic in a population

\item $\frac{dy}{dt} = ky(L-y)$ gives the spread of the disease

\item $L$ is the total population.

\item Use $L = 2.5E5, k = 3E-5, y(0) = 250$

\item Define a function as below

\end{itemize}

\begin{lstlisting}

In []: from scipy.integrate import odeint

In []: def epid(y, t):

  ....     k = 3.0e-5

  ....     L = 2.5e5

  ....     return k*y*(L-y)

  ....

\end{lstlisting}

\end{frame}

\begin{frame}[fragile]

\frametitle{Solving ODEs using SciPy \ldots}

\begin{lstlisting}

In []: t = linspace(0, 12, 61)

In []: y = odeint(epid, 250, t)

In []: plot(t, y)

\end{lstlisting}

%Insert Plot

\end{frame}

\begin{frame}[fragile]

\frametitle{Result}

\begin{center}

\includegraphics[height=2in, interpolate=true]{data/image}  

\end{center}

\end{frame}